hdu 1787(欧拉函数)

GCD Again

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2874 Accepted Submission(s): 1240


[align=left]Problem Description[/align]
Do you have spent some time to think and try to solve those unsolved problem after one ACM contest?
No? Oh, you must do this when you want to become a "Big Cattle".
Now you will find that this problem is so familiar:
The
greatest common divisor GCD (a, b) of two positive integers a and b,
sometimes written (a, b), is the largest divisor common to a and b. For
example, (1, 2) =1, (12, 18) =6. (a, b) can be easily found by the
Euclidean algorithm. Now I am considering a little more difficult
problem:
Given an integer N, please count the number of the integers M (0<M<N) which satisfies (N,M)>1.
This
is a simple version of problem “GCD” which you have done in a contest
recently,so I name this problem “GCD Again”.If you cannot solve it
still,please take a good think about your method of study.
Good Luck!

[align=left]Input[/align]
Input
contains multiple test cases. Each test case contains an integers N
(1<N<100000000). A test case containing 0 terminates the input and
this test case is not to be processed.

[align=left]Output[/align]
For each integers N you should output the number of integers M in one line, and with one line of output for each line in input.

[align=left]Sample Input[/align]

2
4
0

[align=left]Sample Output[/align]

0
1

水题一枚

#include <stdio.h>
#include <string.h>
using namespace std;
typedef long long LL;
LL phi(LL x)
{
LL ans=x;
for(LL i=2; i*i<=x; i++)
if(x%i==0)
{
ans=ans/i*(i-1);
while(x%i==0) x/=i;
}
if(x>1)
ans=ans/x*(x-1);
return ans;
}

int main(){
LL n;
while(scanf("%lld",&n)!=EOF,n){
printf("%lld\n",n-phi(n)-1);
}
}